Jzoj P6357 小ω的图___贪心+并查集

题目大意：

n个点，m条边，问点1走到点n的最大路径and和。
$n<=10^5,m<=5\ast 10^5,边权<2^{63}$

分析：

因为and其实是二进制下的逐位运算，
所以我们可以贪心从二进制的高位开始枚举，
利用并查集判断这一位是否能被加入答案中，
这样显然最优

代码：

#pragma GCC optimize(3)
#include 
#include 
#include 
#include 
#include 
#include 

#define rep(i, st, ed) for (int i = st; i <= ed; i++)
#define rwp(i, ed, st) for (int i = ed; i >= st; i--) 

#define N 100005
#define M 62

using namespace std;

typedef long long ll;

struct Node{
    int u, v; ll w;
}e[N*5];
int f[N], n, m;

void read(int &x) {
	int f = 1; x = 0; char s = getchar(); 
	while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
	while (s >= '0' && s <= '9') { x = x * 10 + (s - '0'); s = getchar(); }
	x = x * f;
}

void read1(ll &x) {
	ll f = 1ll; x = 0ll; char s = getchar();
	while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
	while (s >= '0' && s <= '9') { x = x * 10ll + (s - '0'); s = getchar(); }
	x = x * f;
}

bool cmp(Node aa, Node bb) {
	return aa.w > bb.w;	
}

int Find(int x) {
	return (f[x] == x) ? x : (f[x] = Find(f[x])); 
}

int main() {
	read(n); read(m);
	rep(i, 1, m) read(e[i].u), read(e[i].v), read1(e[i].w);
	sort(e + 1, e + m + 1, cmp);
	int R = n; ll ans = 0;
	rwp(i, M, 0) if ((1ll << i) <= e[n].w) { R = i; break; }
	int xx, yy;
	rwp(k, R, 0) {
		ans |= (1ll << k);
		rep(i, 1, n) f[i] = i;
		rep(i, 1, m) 
		    if (e[i].w >= ans) {
  			    xx = Find(e[i].u);
			    yy = Find(e[i].v);
			    if ((e[i].w & ans) == ans && xx != yy) f[xx] = yy;
		    }
		xx = Find(1); 
		yy = Find(n);
		if (xx != yy) ans ^= (1ll << k);
	}
	printf("%lld\n", ans);
	return 0;
}